What is the solution for y^{''}-3y^'=t^3-6 ?

The solution for y^{''}-3y^'=t^3-6 is y=c_{1}e^{3t}+c_{2}-(t^4}{12}-(t^3)/9-(t^2)/9+\frac{52t)/(27)

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y^{''}-3y^'=t^3-6

y^{^{''}} - 3 y^{^{'}} = t^{3} - 6

y = c_{1} e^{3 t} + c_{2} - \frac{t ^{4}}{12} - \frac{t ^{3}}{9} - \frac{t ^{2}}{9} + \frac{52 t}{27}

Solution steps

y^{''} (t) - 3 y^{'} (t) = t^{3} - 6

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} - \frac{t ^{4}}{12} - \frac{t ^{3}}{9} - \frac{t ^{2}}{9} + \frac{52 t}{27}

y = c_{1} e^{3 t} + c_{2} - \frac{t ^{4}}{12} - \frac{t ^{3}}{9} - \frac{t ^{2}}{9} + \frac{52 t}{27}

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What is the solution for y^{''}-3y^'=t^3-6 ?
The solution for y^{''}-3y^'=t^3-6 is y=c_{1}e^{3t}+c_{2}-(t^4}{12}-(t^3)/9-(t^2)/9+\frac{52t)/(27)